Let α=∑_ k=1 ^ 20 ((π)/(3) k) 4 ^2 ((k π)/(9))-1 and β=∑_ k=1 ^ 20...

$\alpha=\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{3 \pi}{9}+\ldots \ldots+\sin \frac{20 \pi}{9}=\frac{\sin \left(20 \cdot \frac{\pi}{18}\right)}{\sin \frac{\pi}{18}} \cdot \sin \left(\frac{21 \pi}{18}\right)=\cos \frac{\pi}{18}=\cos 10^{\circ}$ \begin{array}{rlrl} \text { IIhy } & \beta & =\sqrt{3} \cos \frac{\pi}{18}=\sqrt{3} \cos 10^{\circ} \\ \beta^2-\alpha^2 & =2 \cos ^2 10^{\circ}=1+\cos 20^{\circ} \equiv 1+\cos \lambda^{\circ} \\ \therefore & \lambda=20 \end{array}